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Calculus/Integral by Parts

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Expert: - 6/12/2010


Question
Hello,
Please help to solve this.

Integral (lnx)^2 dx

Answer
Integration by Parts: INT u dv = uv - INT v du
Let u=ln(x)^2 and dv=dx, thus du=2ln(x)/x dx and v=x
So we get
INT ln(x)^2 dx = xln(x)^2 - INT 2xln(x)/x dx
= xln(x)^2 - 2 INT ln(x) dx

Now integrate by parts again for INT ln(x) dx, which is xln(x)-x+C
So we finally get:
INT ln(x)^2 dx = xln(x)^2 - 2[xln(x) - x] + C
= xln(x)^2 - 2xln(x) + 2x + C

I hope this helps.

Abe

Calculus

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Abe Mantell

Expertise

Hello, I am a college professor of mathematics and regularly teach all levels from elementary mathematics through differential equations, and would be happy to assist anyone with such questions!

Experience

Over 15 years teaching at the college level.

Organizations
NCTM, NYSMATYC, AMATYC, MAA, NYSUT, AFT.

Education/Credentials
B.S. in Mathematics from Rensselaer Polytechnic Institute
M.S. (and A.B.D.) in Applied Mathematics from SUNY @ Stony Brook

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